Upper bound on a sum Some calculations have lead me to the following sum:
$$
\sum_{k=1}^n \left(\frac{1}{n}\right)
^{\left(1\ -\ 2^{\large-k/2}\right)^{\large 2}}.
$$
Simulations and intuition suggests that it is bounded by a constant not depending on $n$.
I would be happy to get some help upper bounding it. 
Thanks.
 A: I felt this must be true, though my comment saying so vanished, for some obscure reason. In the meantime, I managed to make that more precise:
Let $a_k$ be some positive numbers $a_k\le a_1<1.$ We have
$$s_n=\sum^n_{k=1}\left(\frac{1}{n}\right)^{1-a_k}=\frac{1}{n}\sum^n_{k=1}n^{a_k}=\frac{1}{n}\sum^n_{k=1}e^{a_k\ln n}.$$ To estimate that, we use the convexity of $e^x:$ $$e^x\le1+\frac{x}{X}\,(e^X-1)$$ for $x\in[0,X].$ Let $X=a_1\ln n.$ Then, $$e^{a_k\ln n}\le1+\frac{a_k}{a_1}\,(n^{a_1}-1),$$ and by summing up
$$\sum^n_{k=1}e^{a_k\ln n}\le n+\frac{\sum^n_{k=1}a_k}{a_1}\,(n^{a_1}-1)$$ and thus
$$s_n\le1+\frac{\sum^n_{k=1}a_k}{a_1}\,(n^{a_1-1}-n^{-1}).$$ The RHS is clearly bounded, because $a_1-1<0.$ In this special case, we have $a_k=2\cdot2^{-k/2}-2^{-k},$ and our conditions are satisfied ($a_1=\sqrt{2}-1/2<1$). 
A: Let $a_n = 2\ln (2n)/\ln 2.$ Using $(1-2^{-k/2})^2 > 1-2\cdot 2^{-k/2},$ we can see $k\ge a_n$ implies $(1-2^{-k/2})^2 >1-1/n.$ Thus the sum in question is bounded above by
$$\tag 1 \sum_{1\le k \le a_n}\frac{1}{n^{(1-2^{-1/2})^2}} \,+ \sum_{a_n\le k \le n} \frac{1}{n^{1-1/n}}\le \frac{a_n}{n^{(1-2^{-1/2})^2}} + n\cdot \frac{1}{n^{1-1/n}}.$$
Since $a_n$ grows only logarithmically, the first term on the right $\to 0.$ The second term on the right equals $n^{1/n} \to 1.$ Thus the limit of $(1)$ is $1,$ proving your sequence of sums is bounded.
